Orbital shell notation of fluorine is 2. 7 while that of oxygen s 2. 6. This means that these elements (that follow each other in the periodic table) will have high electronegativity in molecules due to their high atomic number (which causes them to strongly attract electron orbital shell closer to their nucleus). NB: Atomic number of a peroid increased from left to right of the periodic table.
Therefore, in the first molecule, the negative dipole would most likely be located between the F atoms In the second molecule the negative molecule would be most likely located in the between the O and F atoms.
Answer:
weak bonds break and strong bonds form
Explanation:
An exothermic reaction is a chemical reaction in which heat energy is evolved during the reaction process.
Bond formation requires energy while bond breakage releases energy. More energy is needed for the formation of weak bonds as compared to strong bonds.
<em>Hence, when weak bonds break, they release more energy than needed to make a corresponding strong bond leading to the release of the remaining energy as heat.</em>
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
Answer:
3.14
Explanation:
A student was comparing two samples with an equal number of carbon atoms. One sample contained only Carbon-12 atoms. One sample contained only Carbon-14 atoms, which contain two more neutrons than Carbon-12 atoms. The student measured the mass of each sample and testing the reactivity of each sample.
Required:
What would best describe the results of the investigation?
Answer:
The total mass of D-Glucose dissolved in a 2μL aliquot is 1 E-4 g
Explanation:
providing a solution to 5% weight-volume as found in commerce:
⇒ % 5 = (5g d-glucose/ 100 mL sln)×100
⇒ 0.05 = g C6H12O6/mL sln
⇒ g C6H12O6 = (2 μL sln)×(0.001 mL/μL)×(0.05 g C6H12O6/mL sln)
⇒ g C6H12O6 = 1 E-4 g C6H12O6
